semaphore concurrency - c

Question

In the following code, wouldn't a mutex be created, in the child, as a copy of its parent's? Hence there're two copies of mutexs now -- one in child, and one in parent. How can it be synchronized? As far as I can remember, you need one copy that are shared by multiple processes in order to make it synchronize.
#include <semaphore.h>
#include <stdio.h>
#include <errno.h>
#include <stdlib.h>
#include <unistd.h>
#include <sys/types.h>
#include <sys/stat.h>
#include <fcntl.h>
#include <sys/mman.h>
int main(int argc, char **argv)
{
int fd, i,count=0,nloop=10,zero=0,*ptr;
sem_t mutex;
//open a file and map it into memory
fd = open("log.txt",O_RDWR|O_CREAT,S_IRWXU);
write(fd,&zero,sizeof(int));
ptr = mmap(NULL,sizeof(int), PROT_READ|PROT_WRITE,MAP_SHARED,fd,0);
close(fd);
/* create, initialize semaphore */
if( sem_init(&mutex,1,1) < 0)
{
perror("semaphore initilization");
exit(0);
}
if (fork() == 0) { /* child process*/
for (i = 0; i < nloop; i++) {
sem_wait(&mutex);
printf("child: %d\n", (*ptr)++);
sem_post(&mutex);
}
exit(0);
}
/* back to parent process */
for (i = 0; i < nloop; i++) {
sem_wait(&mutex);
printf("parent: %d\n", (*ptr)++);
sem_post(&mutex);
}
exit(0);
}

Answer 1

You must not confuse a mutex with a semaphore. A semaphore might allow several threads/processes to access a resource, a mutex allows only ONE concurrent access to a resource.
As stated here you would need to create a named semaphore to make cross-process-synchronisation possible.
You must create a semaphore in your parent process and access is using sem_open in the child process to achieve synchronisation.